Thursday, 25 August 2011

Multiplication at your fingertips!

Here I am going to talk about "Ekadhikena Purveṇa" i.e "One more than the previous one" and very important formula and how to use it for multiplication.

This formula can be applied to multiplication of numbers that satisfy both the following conditions

  1. the same first digit
  2. the sum of their last unit digits is 10.
We have already discussed about applying this formula for deriving squares for any number ending with 5, in previous post (Amazing way to derive squares within seconds). So here, I am going to talk about application of this formula for multiplication of numbers satisfying the conditions stated above.


Let us break the method into two parts:
  1.  Deriving last two digits of answer
  2. Deriving all previous digits of answer

1. Deriving last two digits of answer
These shall be multiplication of last digit of the numbers we are looking to multiply.

2. Deriving all previous digits of answer
All the digits in the multipliers other than the last would constitute base for us and let us call them B1.
Another base is B1 +1 (Because the formula is about one more than the previous one…)
So previous digits in the final answer would be simply [B1 x (B1+1)]


Let us understand this with an example: 24 x 26

Last two digit of the answer:

4 x 6 = 24 (If this answer is in single digit, then we need to put 0 before it)

For other digits, we need B1 and it goes as follows:

B1 = 2 (All the digits other than the last 5 in the number we want to derive square for)
B1 + 1 = 2 + 1 = 3
Therefore other digits of the answer preceding 25 are = 2 x 3 = 6
Hence the final answer would be 624.


So what’s the logical explanation of this wonderful shortcut:
We all know that (a-b) x (a+b) = a2-b2

Now how does this help with multiplication of 24 & 26? It is like this:

24 x 26
(25-1) x (25+1)
252 – 12
625 – 1 (252 derived from the same formula applied for squaring discussed in the previous post)
624 (Well this matches!)

Vedic mathematics rocks!

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